Work And Energy Test

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Work And Energy Test - 7

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following is the biggest unit of energy?

A
Joule

B
Kilowatt hour

C
Erg

D
Electron volt

Solution

1 kWh = 3.6 10⁶ J
1 erg = 1.0 10^–7 J
1 electron volt = 1.6 10^–19 J
Hence, kWh is the largest unit.
Question 2
1 / -0

A person carrying a load of 50 kg on his head moves 10 m on a straight horizontal road. The amount of work done by him is

A
500 joules

B
50_^×9.8 ×10 joules

C
zero

D
None of these

Solution

The amount of work done by a force F, as it moves an object through a distance d, is given by the dot product of force vector and distance vector.
W =
As the force is acting upwards in case of a man lifting a weight on his head and distance moved is in forward direction, the work done is:
W = Fd cos
= Fd cos 90° (angle between F and d is 90°)
= 0
Question 3
1 / -0

A truck and a car are moving on a smooth, level road such that the kinetic energy associated with them is the same. Brakes are applied simultaneously in both of them such that equal retarding forces are produced in both. Which one will cover greater distance before it stops?

A
Car

B
Truck

C
Both will cover the same distance.

D
Nothing can be decided

Solution

Work done by a retarding force is equal to the change in K.E. of a body if the potential energy is constant.
As K.E. for both is same and at the end both stop, therefore work done for the both is same.
Thus,
F d =
As F is also same for both the truck and the car, therefore d is correspondingly same.
Question 4
1 / -0

The work done in lifting 1 kg mass to a height of 9.8 m is about

A
1 J

B
9.8 J

C
(9.8)² J

D
J

Solution

Work done is equal to change in potential energy.
= mgh
= 1 × 9.8 × 9.8 J
= (9.8)² J
Question 5
1 / -0

The kinetic energy of a body becomes four times its initial value while travelling along a straight line. The new momentum will be

A
three times its initial value

B
four times its initial value

C
twice its initial value

D
unchanged

Solution

Initial kinetic energy of the body, KE₁ =

Final kinetic energy of the body, KE₂ =

As per given the condition,

As KE₂ = 4KE₁,

v₂ = 2v₁
Now, momentum is the product of mass and velocity.
Therefore, v₂ = 2v₁ means that new momentum is twice the original momentum.
Question 6
1 / -0

The figure below shows a body of mass 10 kg sliding from rest, down a frictionless track of radius R (in m). Assume that the body started from the top of the track P and slides to the bottom Q. The change in the gravitational potential energy in the motion is (g = acceleration due to gravity)

A
g/2 J

B
10 J

C
10gR J

D
10R J

Solution

The body descends a height R.
The change in the gravitational potential energy would be 'mgh'.
Here, 'm' is the mass of the body, 'g' is the acceleration due to gravity and 'h' is the height.
mgh for the body would be 10gR J.
Question 7
1 / -0

The earth's radius is R and acceleration due to gravity at its surface is g. If a body of mass m is sent to a height h = R/6 from the earth's surface, the potential energy (PE) increases by

A
mgh

B
(7/6) mgh

C
(6/7) mgh

D
(1/7) mgh

Solution

Gravitational potential energy (PE) on surface of earth =
Gravitational PE at a height =
=
=
=
=
= MgR ... (i)
Since, g =
So, (i) becomes △G =
=
= mgh
Question 8
1 / -0

A rifle bullet loses of its velocity in passing through a plank. What is the least number of such planks required just to stop the bullet?

A
5

B
7

C
6

D
8

Solution

Let the thickness of plank be d.
Using the equation of motion,
v² - u² = 2as
The bullet stops finally, so equation can be written as
0 - u² = 2as
Now, let the number of planks required be n, then the equation can be written as
-u² = n2as
u = v_o and s = d
-v_o² = 2nad
n = …… (1)
v = v₀ - (passing through one plank)

-2ad =
-2ad = …… (2)
Putting in (1), n = (Because n has to be a whole number)
Question 9
1 / -0

A particle of mass m is taken along different frictionless paths AB as indicated by the following diagrams. In each case, the particle starts from rest at A and reaches point B. All the paths are in a vertical plane and the external forces act such that the particle moves along the respective paths without restraint.

Which of the following statements will be true in respect of the given diagrams?

A
The kinetic energy of the particle will remain conserved.

B
The potential energy of the particle will remain conserved.

C
The work done in the motion from A to B by the external forces will be the same for all paths.

D
All of the above

Solution

Work done in moving a body by a certain distance against gravity is directly proportional to the height to which the body is moved and is independent of the path followed. Now, in all cases, 'h' is same. Therefore, the work done by the forces in attaining motion from A to B is the same.
Question 10
1 / -0

An object of mass 20 kg falls from rest through a vertical distance of 10 m and acquires a velocity of 10 m/s. The work done by the push of air on the object is (g = 10 m/s²)

A
500 J

B
- 1000 J

C
250 J

D
- 250 J

Solution

Velocity acquired by body = 10 m/s
As body falls 10 m, it`s PE = 20 × 10 × 10 = 2000 J
This is the work done by gravity W_g = 2000 J
Velocity of KE acquired = KE = = 1000 J
The work done by air on an object is actually the work done against motion.

= 1000 J
Question 11
1 / -0

In an inclined plane, we lose in terms of distance but gain in terms of

A
energy

B
work

C
efficiency

D
force

Solution

In an inclined plane the distance travelled by an object is more but the force required is less. Since when work is done energy is conservative, hence, we gain in terms of force.
Question 12
1 / -0

Which of the following options is not a unit of power?

A
Watt

B
Horse power

C
Joule/sec

D
Kilowatt hour

Solution

Kilowatt hour is the unit of energy which is equal to 3.6 10⁶ J.
Question 13
1 / -0

An electric motor creates a tension of 9,000 newton in a hoisting cable and reels it in at the rate of 2 m/s. What is the power of this motor?

A
25 kW

B
18 kW

C
225 watt

D
9000 hp

Solution

Given, F = 9000 N
Also, P = F ×
= 2 m/s and P =
P = kW
Question 14
1 / -0

A man weighing 60 kg lifts a body of 15 kg to the top of a building10 m high in 3 minutes. His efficiency is

A
10%

B
20%

C
30%

D
40%

Solution

Efficiency = %
Question 15
1 / -0

A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time t is proportional to

A
t^1/2

B
t^3/4

C
t^3/2

D
t²

Solution

P = F.v
Power = constant (c)
F.v = cF.s/t = c
m.a.s/t = c
m.a.s = ct
a.s = c/m x t
a.s = bt (c/m = b)
a = v/t
(v/t)s = bt
vs = bt²s/t x s = bt²s² = bt³s = bt^3/2
Question 16
1 / -0

Heat energy is converted into electrical energy by using a

A
hydrometer

B
thermoelectric generator

C
voltmeter

D
photo-cell

Solution

A thermoelectric generator is a solid state device which converts the thermal energy into electrical energy. It is based on the "Seebeck effect".
The Seebeck effect is a phenomenon in which a temperature difference between two dissimilar metals or semiconductors produces a voltage difference between the two substances. When one junction is kept hot and the other junction is kept cold, heated electrons flow towards the cooler one and direct current (DC) flows through that circuit.
The voltages produced by the Seebeck effect are small, usually only a few microvolts per kelvin of temperature difference at the junction.
Question 17
1 / -0

Which of the following options converts light energy into electrical energy?

A
Coolide tube

B
Cloud chamber

C
Photovoltaic cell

D
Thermocouple

Solution

Photovoltaic cell is a semiconductor device which converts the light energy into electric energy. It works based on "photovoltaic effect".
The photovoltaic effect can be defined as being the appearance of a potential difference (voltage) between two layers of a semiconductor slice, in which the conductivities are opposite, or between a semiconductor and a metal, under the effect of a light.
Question 18
1 / -0

A bent bow used for shooting an arrow possesses

A
potential energy

B
kinetic energy

C
momentum

D
velocity

Solution

A bent bow possesses potential energy. This potential energy is then converted to kinetic energy of the arrow.
Question 19
1 / -0

A spring of force constant 60 N/m has an initial stretch 0.2 m. In changing the stretch to 0.25 m, the increase in P.E. is about

A
0.7 joule

B
0.5 joule

C
0.4 joule

D
0.1 joule

Solution

Initial stretch = 0.2 m
P.E. = ^J
=
Now, P.E. = 60 J
=
Charge =
= =
= 0.675 J
or = 0.7 J
Question 20
1 / -0

It requires twenty turns of the stem of a watch to wind the main spring and this stores sufficient energy to keep the watch running for 30.0 hours. If ten turns are given to the stem, it will be

A
20 hr

B
15 hr

C
7.5 hr

D
3.75 hr

Solution

E = 1/2 kx²
For 20 turns E is proportional to (20)²E 400
For 10 turns E is proportional to (10)²
E 100
The energy is one fourth of the first one so secondly it will run for 7.5 hours